if a voltage is applied to a capacitor, current flows easily at first and then slows as the capacitor becomes charged. Inductors
behave just the opposite, in that they reluctantly pass current when a voltage is first applied, and then the current passes easily as time passes. If the input voltage is suddenly raised from zero to some constant value, sketch the current in the capacitor, iC, and the inductor, iL, as a function of time. What is the steady-state current in the capacitor and inductor?
1 answer:
Answer: photo!
Explanation:
For the capacitor, the current as a function of time will be
I(t)=V/R e^(-t/RC)
In steady-state, time tends to infinite, so current tends to zero.
For the inductor, the current as a function of time is
I(t)=V/R (1-e^(-tR/L))
In steady-state, time tends to infinite, so current tends to V/R
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Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)
Now we can use the equation that defines the percentage of increase, in this case for speed
Now we use the equation obtained in the previous step, and replace values
the percent increase in the velocity of air is 25.65%